What is the value of the following logarithm? $\log_{9} \left(\dfrac{1}{81}\right)$
Solution: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $9^{y} = \dfrac{1}{81}$ In this case, $9^{-2} = \dfrac{1}{81}$, so $\log_{9} \left(\dfrac{1}{81}\right) = -2$.